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Bài 2:
a: \(=\dfrac{7}{9}\left(\dfrac{7}{6}-\dfrac{19}{20}-\dfrac{1}{15}\right)+\dfrac{22}{5}\cdot\dfrac{1}{24}\)
\(=\dfrac{7}{9}\cdot\dfrac{3}{20}+\dfrac{22}{120}=\dfrac{7}{60}+\dfrac{11}{60}=\dfrac{18}{60}=\dfrac{3}{10}\)
b: \(=\left(\dfrac{35-32}{60}\right)^2+\dfrac{4}{5}\cdot\dfrac{70-45}{80}\)
\(=\dfrac{1}{400}+\dfrac{4\cdot25}{400}=\dfrac{101}{400}\)
a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729
b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)
c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)
\(=\dfrac{2^3}{4^4}\cdot7\)
\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)
d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)
\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)
\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)
e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)
\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)
\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)
\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)
f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)
\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)
\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)
g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)
\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)
\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)
\(\dfrac{3!.4.5.\left(m-1\right)!.m.\left(m+1\right)}{m\left(m+1\right)3!.\left(m-1\right)!}\)
=\(4.5\)
=20
a: \(\sqrt[4]{\left(-\dfrac{4}{5}\right)^4}=\left|-\dfrac{4}{5}\right|=\dfrac{4}{5}\)
b: \(\dfrac{\sqrt{4}}{\sqrt{5}}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)
c: \(\left(\sqrt[3]{9}\right)^2=\left(9^{\dfrac{1}{3}}\right)^2=9^{\dfrac{2}{3}}\)
d: \(\sqrt[5]{\sqrt{a}}=\sqrt[5]{a^{\dfrac{1}{2}}}=a^{\dfrac{1}{2}\cdot\dfrac{1}{5}}=a^{\dfrac{1}{10}}\)
e: \(\sqrt[3]{2^6}=\sqrt[3]{\left(2^2\right)^3}=2^2=4\)
a)ĐKXĐ:\(a\ge0;a\ne16\)
\(B=\left[\dfrac{3\sqrt{a}}{\sqrt{a}+4}+\dfrac{\sqrt{a}}{\sqrt{a}-4}+\dfrac{4\left(a+2\right)}{16-a}\right]:\left(1-\dfrac{2\sqrt{a}+5}{\sqrt{a}+4}\right)\)
=\(\dfrac{3\sqrt{a}\left(\sqrt{a}-4\right)+\sqrt{a}\left(\sqrt{a}+4\right)-4\left(a+2\right)}{a-16}:\dfrac{\sqrt{a}+4-2\sqrt{a}-5}{\sqrt{a}+4}=\dfrac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\cdot\dfrac{\sqrt{a}+4}{-\sqrt{a}-1}=\dfrac{-8\sqrt{a}-8}{\left(\sqrt{a}-4\right)\left(-\sqrt{a}-1\right)}=\dfrac{8\left(-\sqrt{a}-1\right)}{\left(\sqrt{a}-4\right)\left(-\sqrt{a}-1\right)}=\dfrac{8}{\sqrt{a}-4}\)
Vậy...
b)Với \(a\ge0;a\ne16\) thì B=\(\dfrac{8}{\sqrt{a}-4}\)
B=-3 thì \(\dfrac{8}{\sqrt{a}-4}=-3\)
=>\(9=-3\sqrt{a}+24\)
<=>-15=-3\(\sqrt{a}\)
<=>\(\sqrt{a}=5\)
<=>a=25(TM)
Vậy a=25 thì B=-3
c)Với \(a\ge0;a\ne16\) thì B=\(\dfrac{8}{\sqrt{a}-4}\)
| \(\sqrt{a}-4\) | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
| \(\sqrt{a}\) | -4(L) | 0 | 2 | 3 | 5 | 6 | 8 | 12 |
| \(\sqrt{a}\) | 0 | 2 | 3 | 5 | 6 | 8 | 12 |
| a | 0(TM) | 4(TM) | 9(TM) | 25(TM) | 36(TM) | 64(TM) | 144(TM) |
(BẠN KẺ 1 BẢNG 3 HÀNG THÔI NHA,MÌNH KẺ LỖI NÊN LÀM 2 BẢNG)
Vậy...
\(E=a^{12-4}.b^{3-7}=\dfrac{a^8}{b^4}\)
\(E=a^{4-6}.b^{3.4}=\dfrac{b^{12}}{a^2}\)
\(F=\dfrac{a^{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}}{a^{\left(\sqrt{5}-3\right)+\left(4-\sqrt{5}\right)}}=\dfrac{a^2}{a^1}=a\)
\(=\dfrac{7!4!8!}{10!3!5!}-\dfrac{7!4!9!}{10!2!7!}=\left(\dfrac{4!}{3!}\right)\left(\dfrac{8!}{10!}\right)\left(\dfrac{7!}{5!}\right)-\left(\dfrac{7!}{7!}\right)\left(\dfrac{4!}{2!}\right)\left(\dfrac{9!}{10!}\right)\)
\(=4.\left(\dfrac{1}{10.9}\right).7.6-1.\left(4.3\right).\left(\dfrac{1}{10}\right)=...\)